我从mysql获取数据时收到数据库错误[重复]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

I am developing a e-commerce site . I want to get the username $_SESSION['REMOTE_ADDR']

the code is here :

$new_sql    = 'SELECT
               sum(amount) as items_total
               FROM
               products_added
               where `username` = '.mysqli_real_escape_string($conn, $_SESSION["REMOTE_ADDR"]).'
               ORDER BY id';

        $resu       =  mysqli_query($conn, $new_sql);
        $itemsTotal =  mysqli_fetch_array($resu);
        $grandTotal =  ($itemsTotal['items_total']);
        echo $grandTotal;

But getting:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\orlando\php\include\content.php on line 120

</div>

Answer 1

Use the following method. Escape the variable before, it's easier.

Sidenote:

I'm pretty sure you meant to use $_SERVER["REMOTE_ADDR"] instead of $_SESSION["REMOTE_ADDR"].

$_SESSION["REMOTE_ADDR"] isn't a valid session variable.

Consult: http://php.net/manual/en/reserved.variables.server.php on $_SERVER variable(s).

Code:

$remote_address = mysqli_real_escape_string($conn, $_SERVER["REMOTE_ADDR"]);
$new_sql    = "SELECT
               sum(amount) as items_total
               FROM
               products_added
               where `username` = '".$remote_address."'
               ORDER BY id";

$resu = mysqli_query($conn, $new_sql) or die(mysqli_error($conn));
$itemsTotal =  mysqli_fetch_array($resu);
$grandTotal =  $itemsTotal['items_total'];
echo $grandTotal;

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Footnotes:

However, and as Gordon (Linoff) states in a comment:

You have a sum() with no group by, so it is returning one row. However, you have an order by suggesting that you are expecting more than one row.